Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 13J (Statistical Modelling)

Table of contents
- a 13J
  - Solution a
- b 13J
  - Solution b
- c 13J
  - Solution c

a

Solution

S_{i} (t) = 1 - F_{i} (t)

for the survival function, the definition of the hazard function gives

h_{i} (t) = \frac{f _{i} ( t )}{S _{i} ( t )} = - \frac{S _{i}^{'} ( t )}{S _{i} ( t )} = - \frac{d}{d t} lo g S_{i} (t) .

Since

S_{i} (0) = 1

, integration yields

S_{i} (t) = exp [- \int_{0}^{t} h_{i} (s) d s] .

Therefore

F_{i} (t) = 1 - exp [- \int_{0}^{t} h_{i} (s) d s] .

The exponent is the negative cumulative hazard.

b

Solution

Let

Λ (t) = \int_{0}^{t} λ (s) d s

be the baseline cumulative hazard. Under the proportional hazards model,

h_{i} (t) = λ (t) e^{β^{T} x_{i}}, S_{i} (t) = exp [- e^{β^{T} x_{i}} Λ (t)] .

Thus

f_{i} (Y_{i}) = h_{i} (Y_{i}) S_{i} (Y_{i}) = λ (Y_{i}) e^{β^{T} x_{i}} exp [- e^{β^{T} x_{i}} Λ (Y_{i})] .

Independence makes the log likelihood

ℓ (β) = i = 1 \sum n [lo g λ (Y_{i}) + β^{T} x_{i} - e^{β^{T} x_{i}} Λ (Y_{i})] .

c

Solution

Fit a Poisson regression to artificial responses

Z_{i} = 1

with means

μ_{i} = Λ (Y_{i}) e^{β^{T} x_{i}} .

With the logarithmic link, this is the generalized linear model

lo g μ_{i} = β^{T} x_{i} + lo g Λ (Y_{i}),

where

lo g Λ (Y_{i})

is a known offset. Its log likelihood, apart from constants, is

ℓ_{P} (β) = \sum_{i} [lo g Λ (Y_{i}) + β^{T} x_{i} - Λ (Y_{i}) e^{β^{T} x_{i}}] .

ℓ (β) - ℓ_{P} (β) = \sum_{i} [lo g λ (Y_{i}) - lo g Λ (Y_{i})]

does not depend on

β

. The Poisson surrogate therefore has exactly the same maximum-likelihood estimate of $β$ . This is the Poisson surrogate for an uncensored proportional-hazards likelihood.

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